Chapter 1.2, Problem 21E

To determine

To calculate:the values for the product of the functions $f\left(x\right)=2x+3$ and $g\left(x\right)=3x-5$ , at $x=-2,x=-1,x=0,x=1$ and $x=2$ , graph component piece functions and their product.

Answer to Problem 21E

The value of the functions and product of the functions for different values of x are,

  $\begin{array}{cccc}x& f\left(x\right)& g\left(x\right)& \left(f\cdot g\right)\left(x\right)\\ -2& -1& -11& 11\\ -1& 1& -8& -8\\ 0& 3& -5& -15\\ 1& 5& -2& -10\\ 2& 7& 1& 7\end{array}$

The graph of the functions $f\left(x\right)=2x+3$ , $g\left(x\right)=3x-5$ and $\left(f\cdot g\right)\left(x\right)=6x^2-x-15$ is provided below,

  

Explanation of Solution

Given information:

The functions, $f\left(x\right)=2x+3$ and $g\left(x\right)=3x-5$ .

Formula used:

The multiplication or product of the functions $f$ and $g$ is represented as $f\cdot g$ , mathematically it is denoted by,

  $\left(f\cdot g\right)\left(x\right)=f\left(x\right)\cdot g\left(x\right)$

Calculation:

Consider the provided functions, $f\left(x\right)=2x+3$ and $g\left(x\right)=3x-5$ .

Recall the multiplication or product of the functions $f$ and $g$ is represented as $f\cdot g$ , mathematically it is denoted by,

  $\left(f\cdot g\right)\left(x\right)=f\left(x\right)\cdot g\left(x\right)$

Evaluate the value of $f\left(x\right)\cdot g\left(x\right)$ that is product of the functions,

  $f\left(x\right)\cdot g\left(x\right)=\left(2x+3\right)\cdot \left(3x-5\right)$

Multiply the first term of the first bracket with the terms in second bracket and do the same with the second term of the first bracket.

  $\begin{array}{c}f\left(x\right)\cdot g\left(x\right)=\left(2x+3\right)\cdot \left(3x-5\right)\\ =2x\left(3x-5\right)+3\left(3x-5\right)\\ =6x^2-10x+9x-15\end{array}$

Combine the like terms,

  $\begin{array}{c}f\left(x\right)\cdot g\left(x\right)=\left(2x+3\right)\cdot \left(3x-5\right)\\ =2x\left(3x-5\right)+3\left(3x-5\right)\\ =6x^2-10x+9x-15\\ =6x^2-x-15\end{array}$

Therefore, the value of product of the function is $6x^2-x-15$ .

Now, evaluate the value of the individual function $f\left(x\right)=2x+3$ , $g\left(x\right)=3x-5$ and product of the function, $f\left(x\right)\cdot g\left(x\right)=6x^2-x-15$ at the provided points $x=-2,x=-1,x=0,x=1$ and $x=2$ .

For $x=-2$ , substitute the value of x in the functions, $f\left(x\right)=2x+3$ , $g\left(x\right)=3x-5$ and $f\left(x\right)\cdot g\left(x\right)=6x^2-x-15$ .

  $\begin{array}{c}f\left(-2\right)=2\left(-2\right)+3\\ =-4+3\\ =-1\end{array}$

Next,

  $\begin{array}{c}g\left(-2\right)=3\left(-2\right)-5\\ =-6-5\\ =-11\end{array}$

Last,

  $\begin{array}{c}f\left(-2\right)\cdot g\left(-2\right)=6{\left(-2\right)}^2-\left(-2\right)-15\\ =24+2-15\\ =11\end{array}$

For $x=-1$ , substitute the value of x in the functions, $f\left(x\right)=2x+3$ , $g\left(x\right)=3x-5$ and $f\left(x\right)\cdot g\left(x\right)=6x^2-x-15$ .

  $\begin{array}{c}f\left(-1\right)=2\left(-1\right)+3\\ =-2+3\\ =1\end{array}$

Next,

  $\begin{array}{c}g\left(-1\right)=3\left(-1\right)-5\\ =-3-5\\ =-8\end{array}$

Last,

  $\begin{array}{c}f\left(-1\right)\cdot g\left(-1\right)=6{\left(-1\right)}^2-\left(-1\right)-15\\ =6+1-15\\ =-8\end{array}$

For $x=0$ , substitute the value of x in the functions, $f\left(x\right)=2x+3$ , $g\left(x\right)=3x-5$ and $f\left(x\right)\cdot g\left(x\right)=6x^2-x-15$ .

  $\begin{array}{c}f\left(0\right)=2\left(0\right)+3\\ =3\end{array}$

Next,

  $\begin{array}{c}g\left(0\right)=3\left(0\right)-5\\ =-5\end{array}$

Last,

  $\begin{array}{c}f\left(0\right)\cdot g\left(0\right)=6{\left(0\right)}^2-\left(0\right)-15\\ =-15\end{array}$

For $x=1$ , substitute the value of x in the functions, $f\left(x\right)=2x+3$ , $g\left(x\right)=3x-5$ and $f\left(x\right)\cdot g\left(x\right)=6x^2-x-15$ .

  $\begin{array}{c}f\left(1\right)=2\left(1\right)+3\\ =2+3\\ =5\end{array}$

Next,

  $\begin{array}{c}g\left(1\right)=3\left(1\right)-5\\ =3-5\\ =-2\end{array}$

Last,

  $\begin{array}{c}f\left(1\right)\cdot g\left(1\right)=6{\left(1\right)}^2-\left(1\right)-15\\ =6-1-15\\ =-10\end{array}$

For $x=2$ , substitute the value of x in the functions, $f\left(x\right)=2x+3$ , $g\left(x\right)=3x-5$ and $f\left(x\right)\cdot g\left(x\right)=6x^2-x-15$ .

  $\begin{array}{c}f\left(2\right)=2\left(2\right)+3\\ =4+3\\ =7\end{array}$

Next,

  $\begin{array}{c}g\left(2\right)=3\left(2\right)-5\\ =6-5\\ =1\end{array}$

Last,

  $\begin{array}{c}f\left(2\right)\cdot g\left(2\right)=6{\left(2\right)}^2-\left(2\right)-15\\ =24-2-15\\ =7\end{array}$

Thus, the above results are tabulated as,

  $\begin{array}{cccc}x& f\left(x\right)& g\left(x\right)& \left(f\cdot g\right)\left(x\right)\\ -2& -1& -11& 11\\ -1& 1& -8& -8\\ 0& 3& -5& -15\\ 1& 5& -2& -10\\ 2& 7& 1& 7\end{array}$

Plot the graph of the functions $f\left(x\right)=2x+3$ , $g\left(x\right)=3x-5$ and $\left(f\cdot g\right)\left(x\right)=6x^2-x-15$ on the same coordinate axes and visualize the each of them.

  

Thus, the results obtained graphically and mathematically are same.